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NYT Pips answers — Tuesday, April 21, 2026

Pips #247 · Edited by Ian Livengood · Easy by Ian Livengood · Medium by Ian Livengood · Hard by Rodolfo Kurchan

Stuck on today's New York Times Pips? Below are three progressive hints for each difficulty, then the full solution as a tap-to-reveal board. Hints first — the answer only when you want it.

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Easy Pips answer — April 21, 2026

5×4 board · 6 dominoes · 5 regions · by Ian Livengood

Easy today is a 5×4 board with 6 dominoes and 5 regions. The mix is 3 sum regions, 1 "less than", 1 free region. Three cells adding to 17 at row 1, column 1 needs big pips — at least one 6.. Rack check: 2 doubles, 4 blank halves, 2 sixes.

Hints

Hint 1 — where to start

Three cells adding to 17 at row 1, column 1 needs big pips — at least one 6.

Hint 2 — first domino

Place the 6–5 domino with its 6 at row 1, column 1 and its 5 at row 2, column 1.

Hint 3 — second domino

Next, the 5–0 goes at row 5, column 4 → row 4, column 4. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: 17 must add up to 17; 9 must add up to 9; 0 must add up to 0; <6 must add up to less than 6.

17
9
0
<6
Solution as text
  1. 4–4 — 4 at row 1, column 2, 4 at row 1, column 3
  2. 0–6 — 0 at row 4, column 1, 6 at row 3, column 1
  3. 0–0 — 0 at row 4, column 2, 0 at row 4, column 3
  4. 1–2 — 1 at row 1, column 4, 2 at row 2, column 4
  5. 6–5 — 6 at row 1, column 1, 5 at row 2, column 1
  6. 5–0 — 5 at row 5, column 4, 0 at row 4, column 4

Medium Pips answer — April 21, 2026

5×6 board · 9 dominoes · 9 regions · by Ian Livengood

9 dominoes, 9 regions, a 5×6 grid — that's the medium puzzle for April 21, 2026. The mix is 4 sum regions, 3 "all equal", 1 "less than", 1 free region. Best foothold — the single cell at row 3, column 3 is marked 1, so it can only hold a 1. After that, the "=" region starting at row 3, column 4 has 3 cells that must all match — every 6 in the puzzle should come here.. Rack check: 2 doubles, 3 blank halves, 3 sixes.

Hints

Hint 1 — where to start

The single cell at row 3, column 3 is marked 1, so it can only hold a 1. Find the domino with a 1 that fits there.

Hint 2 — first domino

Place the 6–1 domino with its 6 at row 3, column 4 and its 1 at row 3, column 3.

Hint 3 — second domino

Next, the 2–0 goes at row 1, column 5 → row 1, column 6. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: = must all be the same number; 8 must add up to 8; <2 must add up to less than 2; 1 must add up to 1; 5 must add up to 5.

=
8
<2
8
1
=
5
=
Solution as text
  1. 4–2 — 4 at row 4, column 1, 2 at row 5, column 1
  2. 6–1 — 6 at row 3, column 4, 1 at row 3, column 3
  3. 2–2 — 2 at row 1, column 3, 2 at row 1, column 4
  4. 0–1 — 0 at row 5, column 4, 1 at row 5, column 5
  5. 3–0 — 3 at row 5, column 2, 0 at row 5, column 3
  6. 6–6 — 6 at row 3, column 5, 6 at row 4, column 5
  7. 2–5 — 2 at row 1, column 2, 5 at row 1, column 1
  8. 4–5 — 4 at row 3, column 1, 5 at row 2, column 1
  9. 2–0 — 2 at row 1, column 5, 0 at row 1, column 6

Hard Pips answer — April 21, 2026

7×5 board · 13 dominoes · 12 regions · by Rodolfo Kurchan

The hard board runs 7 rows by 5 columns: 13 dominoes to place across 12 regions. You're working with 9 sum regions, 1 "all equal", 1 "less than". The way in: the single cell at row 2, column 1 is marked 4, so it can only hold a 4. After that, the single cell at row 2, column 3 is marked 6, so it can only hold a 6. Rack check: 4 doubles, 2 blank halves, 4 sixes.

Hints

Hint 1 — where to start

The single cell at row 2, column 1 is marked 4, so it can only hold a 4. Find the domino with a 4 that fits there.

Hint 2 — first domino

Place the 4–4 domino with its 4 at row 2, column 1 and its 4 at row 3, column 1.

Hint 3 — second domino

Next, the 6–6 goes at row 2, column 3 → row 3, column 3. From there the rest should fall into place.

Solution

Tap any cell to reveal its pips, or reveal the whole board. Region rules: = must all be the same number; 2 must add up to 2; 4 must add up to 4; ≠ must all be different; 6 must add up to 6.

=
2
4
≠
6
0
12
<2
10
12
9
6
Solution as text
  1. 4–4 — 4 at row 2, column 1, 4 at row 3, column 1
  2. 3–5 — 3 at row 7, column 3, 5 at row 7, column 2
  3. 4–1 — 4 at row 1, column 3, 1 at row 1, column 4
  4. 5–5 — 5 at row 4, column 1, 5 at row 5, column 1
  5. 4–0 — 4 at row 1, column 2, 0 at row 2, column 2
  6. 1–1 — 1 at row 3, column 2, 1 at row 4, column 2
  7. 4–5 — 4 at row 7, column 1, 5 at row 6, column 1
  8. 6–6 — 6 at row 2, column 3, 6 at row 3, column 3
  9. 4–3 — 4 at row 4, column 5, 3 at row 3, column 5
  10. 5–6 — 5 at row 5, column 5, 6 at row 6, column 5
  11. 2–3 — 2 at row 4, column 3, 3 at row 5, column 3
  12. 6–3 — 6 at row 7, column 5, 3 at row 7, column 4
  13. 1–0 — 1 at row 1, column 5, 0 at row 2, column 5

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